Let $\tau>0$, $d\in\mathbb N$, $v:[0,\tau]\times\mathbb R^d\to\mathbb R^d$ with $v\in C^{0,\:1}(\mathbb R^d,C^0([0,\tau],\mathbb R^d))$ and $T:[0,\tau]\times\mathbb R^d\to\mathbb R^d$ be continuously differentiable in the first argument with $$\frac{\partial T}{\partial t}(t,x)=v(t,T_t(x))\;\;\;\text{for all }(t,x)\in(0,\tau)\times\mathbb R^d\tag1$$ and $$T_0=\operatorname{id}_{\mathbb R^d}.\tag2$$
Are we able to show that $T\in C^{0,\:1}(\mathbb R^d,C^1([0,\tau],\mathbb R^d))$, i.e. $$\max\left(\sup_{t\in[0,\:\tau]}\left\|T_t(x)-T_t(y)\right\|,\sup_{t\in[0,\:\tau]}\left\|\frac{\partial T}{\partial t}(t,x)-\frac{\partial T}{\partial t}(t,y)\right\|\right)\le c\left\|x-y\right\|\tag3$$ for all $x,y\in\mathbb R^d$ for some $c\ge0$?
Note that \begin{equation}\begin{split}\left\|T_t(x)-T_t(y)\right\|&=\left\|\int_0^tv(s,T_s(x))-v(s,T_s(y))\:{\rm d}s+x-y\right\|\\&\le\int_0^t\left\|v(s,T_s(x))-v(s,T_s(y))\right\|+\left\|x-y\right\|\end{split}\tag4\end{equation} for all $t\in[0,\tau]$ and $x,y\in\mathbb R^d$ and, if $c_1$ is the Lipschitz constant of $v$, $$\left\|v(s,T_s(x))-v(s,T_s(y))\right\|+\left\|x-y\right\|\le c_1\left\|T_s(x)-T_s(y)\right\|\tag5$$ for all $s\in[0,\tau]$ and $x,y\in\mathbb R^d$.
EDIT: The desired claim can be found in Theorem 4.1 of Shapes and Geometries: Metrics, Analysis, Differential Calculus, and Optimization, Second Edition:
You need to treat the two parts of the integral formula separately to get $$ \|T_t(x)-T_t(y)\|\le\|x-y\|+c\int_0^t\|T_s(x)-T_s(y)\|\,ds $$ This now is the standard situation for the Grönwall lemma giving $$ \|T_t(x)-T_t(y)\|\le e^{ct}\|x-y\| $$ so that the claim is true with $c_1=e^{c\tau}$.
That the exponential in the new Lipschitz constant can not be avoided shows the standard example of the vector field $v(t,x)=cx$ where then $T_t(x)=xe^{ct}$.
Insertion of that result then shows that $$ \|\partial_tT_t(x)-\partial_tT_t(y)\| =\|v(t,T_t(x))-v(t,T_t(y)) \le c\|T_t(x)-T_t(y)\|\le ce^{ct}\|x-y\| $$